Simple Harmonic Motion and Uniform Circular Motion — Question 6

Question 6

Harmonic coordinates do not guarantee circular motion

A particle moves in a plane with $x=A\cos\omega t$ and $y=B\sin\omega t$, where $A,B>0$. Both coordinates are harmonic.

1. Find the path and derive the speed as a function of phase.

2. For $A=2B$, compare speeds at the positive $x$ and $y$ extremes. Identify the condition for uniform circular motion.

Question 6 - Solution

Harmonic coordinates do not guarantee circular motion

(a) Derivation and calculation

$$\frac{x^2}{A^2}+\frac{y^2}{B^2}=\cos^2\omega t+\sin^2\omega t=\boxed1.$$$$\mathbf v=(-A\omega\sin\omega t,\ B\omega\cos\omega t),$$$$\boxed{v=\omega\sqrt{A^2\sin^2\omega t+B^2\cos^2\omega t}}.$$

(b) Result and interpretation

$$v(A,0)=B\omega,\quad v(0,B)=A\omega.$$$$A=2B\Rightarrow\boxed{v(0,B)/v(A,0)=2}.$$$$A=B\Rightarrow r=A,\quad v=A\omega=\text{constant}.$$Thus equal amplitudes are required for uniform circular motion in this quadrature family, even though $\mathbf a=-\omega^2\mathbf r$ in every case.

Physical check

Central acceleration alone permits elliptical as well as circular motion.